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Unit 7 Notes

Hidaya’s Story

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Overview of Past

Genetic Information

Humans have 46 chromosomes = 23 homologous pairs (each pair contains 1 chromosome inherited from mom and 1 chromosome inherited from dad)

Chromosomes are made of DNA → DNA contains genes which code for proteins which in turn create physical characteristics (phenotypes)

Is this individual biologically a male or a female? How do you know?

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Pedigrees

A family tree that tracks a particular trait.

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Practice

In the Martinez family, tongue rolling is dominant (R). Grandpa Javier was a roller (Rr) and Grandma Rosa was a non-roller (rr). Their children were Miguel (Rr, roller) and Ana (rr, non-roller). Miguel married Elena (rr) and all three of their children were rollers (Rr). Ana married Luis (Rr) and they had one roller child (Rr) and one non-roller (rr). Students can use this to create a simple pedigree.

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Pedigrees (cont.)

Traits can be dominant (Genotype = AA or Aa) or recessive (Genotype = aa).

→ Dominant Examples = Freckles and Dimples

→ Recessive Examples = Hitchhiker’s Thumb and Attached Earlobes

Freckles

Hitchhiker’s Thumb

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Albinism (We will first focus on a Recessive Trait)

Trait: Skin Color

Phenotype: Normal Skin Pigment (dominant) vs. Albino Skin (recessive)

Genotype: *Remember genotypes are made up of TWO ALLELES (1 from each parent)

  • AA and Aa code for normal skin pigmentation
    • Aa is considered a carrier of the trait = these individuals do not have the disease but can pass the disease to future children
  • aa codes for albino skin

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Practice Punnett Square

Mother’s Genotype: Aa

Father’s Genotype: Aa

Genotype %:

AA 25%

Aa 50%

aa 25%

Phenotype %:

Normal Skin Pigment 75%

Albino Skin 25%

Genotype = 2 letter combination → Must write out all even if the % = 0

Phenotype = physical characteristic → Again you must write out all even if the % = 0

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Now you try TWO different crosses →

Aa x aa

Aa x AA

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AUTOSOMAL TRAITS

Autosomal traits can be DOMINANT (AA or Aa) or RECESSIVE (aa)

→ We have covered an autosomal recessive trait called albinism

Autosomal Traits: the gene of the trait is carried on an autosomal chromosome (any chromosome #1-22 = does not include the sex chromosomes #23)

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SEX-LINKED TRAITS

Sex-Linked Traits: the gene of the trait is either carried on the X or Y chromosome (sex chromosomes are pair #23)

  • Males and females do not have the same sex chromosomes which causes a difference in how these sex-linked traits are expressed
    • Use XX for females and XY for males

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  • The letters that represent the trait being studied will correspond to the name of the trait in some way
    • Those letters are written as superscripts (XBXb vs. XBY)

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Huntington’s Disease (Dominant Trait)

Trait: Huntington’s Disease

Phenotype: Huntington’s (dominant) vs. No Huntington’s (recessive)

Genotype: *Remember genotypes are made up of TWO ALLELES (1 from each parent)

  • AA and Aa code for Huntington’s Disease
  • aa codes for NO Huntington’s Disease

*NO carriers → you either have the disease or don’t

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Color Blindness (X-linked Recessive Trait)

Trait: Color Vision

Phenotype: Normal Vision (dominant) vs. Color Blindness (recessive)

Genotype: *This one is SPECIAL → Males only inherit 1 recessive allele vs. females must inherit 2 recessive alleles (*Females can be carriers)

Males:

XBY - normal vision

XbY - color blindness

Females:

XBXB - normal vision

XBXb - normal vision *Carrier of the trait

XbXb - color blindness

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Fragile X Syndrome (X-linked Dominant Trait)

Trait: Fragile X Syndrome

Phenotype: Fragile X Syndrome or No Disease

Genotype: *This one is SPECIAL → Males only inherit 1 dominant allele vs. females can inherit 1 or 2 dominant alleles *NO carriers → you either have the disease or don’t

Males:

XBY - Fragile X

XbY - No disease

Females:

XBXB - Fragile X

XBXb - Fragile X

XbXb - No disease

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Practice Problem for SEP Assessment

Maria and James are a married couple who both have freckles. They have three children: Alex, who has freckles, and twins, Sam and Riley, who do not have freckles.

Maria's parents both had freckles, while James' mother had freckles, but his father did not have freckles. Maria's brother, David, does not have freckles, even though both of their parents have freckles.

Freckles (F) is dominant, and no freckles (f) is recessive.

  1. Draw a pedigree displaying the entire family. Which two individuals could be FF or Ff??
  2. If Alex is heterozygous for freckles and marries someone who is also heterozygous, what are the chances they have a child with no freckles? Set up a punnett square to support your answer.

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The Father of Genetics

  • Gregor Mendel
    • Studied the different traits of pea plants

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Non-Mendelian Genetics

  • Mendel studied genes of single traits which only consisted of 2 alleles where the capital allele had complete dominance over the lowercase allele.
    • We will determine that this is NOT ALWAYS TRUE!

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Incomplete Dominance

  • Incomplete Dominance is a form of intermediate inheritance in which one allele for a specific trait is not completely expressed over its paired allele.
  • The results show a third phenotype, which is a combination of both phenotypes
    • Similar to mixing paint colors

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Common Examples of Incomplete Dominance

  1. Flowers: Red x white = pink (specific species - see picture)
  2. Red bird x blue bird = purple birds (specific species)
  3. Curly hair x straight hair = wavy hair
  4. Black sheep x white sheep = grey sheep

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Incomplete Dominance Practice Problem

In humans, hair type is determined by incomplete dominance. Some people have curly hair (CC), some people have straight hair (SS), and heterozygous have wavy hair (CS). What would occur if a person with wavy hair had a baby with a person with straight hair?

  1. Write parent genotypes.
  2. Set up and complete a punnett square.
  3. Write the genotype and phenotype percentages/fractions.

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Codominance

  • Codominance: is a form of dominance where the alleles of a heterozygous genotype are BOTH fully expressed in the phenotype.
  • This results in offspring with a phenotype that is neither dominant nor recessive, but both phenotypes are shown

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Common Codominant Examples

  1. Flowers: White x red = red and white flowers (specific species - see picture)
  2. Chickens: black x white = checkered black and white
  3. Cows: red x white = roan (red and white hairs)
  4. Blood Type: A x B = AB

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Codominant Practice Problem

Cat fur is determined by codominance. The genotypes are: tan fur (TT), black fur (BB), and tan and black spots (TB - called a ‘tabby cat’). What would occur if two tabby cats were crossed?

  1. Write parent genotypes.
  2. Set up and complete a punnett square.
  3. Write the genotype and phenotype percentages/fractions.

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Summarize

https://www.youtube.com/watch?v=VG0Ktdpj7Rg

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Multiple ALleles - Blood Type

  • Blood types consist of 3 possible alleles instead of just 2 like Mendel studied
    • IA = dominant allele
    • IB = dominant allele
    • i = recessive allele
  • Blood types are named based on the protein attached to the outside of red blood cells
  • Blood type AB also shows codominance

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Possible Genotypes for all Blood Types

Homozygous Dominant

Homozygous Recessive

Heterozygous

Codominant

Phenotype is the Blood Type

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Write the correct GENOTYPE for each person:

  1. Heterozygous Blood Type A
  2. Blood Type AB
  3. Homozygous Blood Type B
  4. Blood Type O
  5. Blood Type B for a person whose mother is Blood Type O
  6. Blood Type A for a person whose father is homozygous for Blood Type A and their mother is Blood Type AB
  1. IAi
  2. IAIB
  3. IBIB
  4. ii
  5. IBi
  6. IAIA

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Blood Type Practice Problems

A man with type A blood, whose father was type B marries a woman with type AB blood. Set up a punnett square and list the genotypes and phenotypes of their children.

  1. Genotypes (percent/fractions):
  2. Phenotypes (percent/fractions):

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Blood Summary Video

https://www.youtube.com/watch?v=KXTF7WehgM8

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Information about the

Blood Typing Lab

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A Blood

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B Blood

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AB Blood

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O Blood

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The next set of slides is for Honors Biology Only

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Dihybrid Cross

  • Dihybrid Cross: is a cross of 2 traits/genes = 4 total alleles; 2 alleles per gamete

Trait

Dominant

Recessive

Flower Color

Purple (P)

White (p)

Plant Height

Tall (T)

Short (t)

Seed Color

Yellow (Y)

Green (y)

Pod Shape

Round (R)

Wrinkled (r)

Write the correct genotype for each:

  1. Homozygous purple flowers, homozygous green seed
  2. Heterozygous tall plant height, homozygous round pod shape
  3. Heterozygous yellow seed and purple flowers
  4. Short plant with wrinkled pods
  5. White flowers with heterozygous round pod shape

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Dihybrid Cross

  • Use FOIL

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TtSS

Possible Gametes:

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TS, TS, tS, tS

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*** Do not write duplicate gametes on the dihybrid punnett square.

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Find all possible gametes for each.

  1. PPyy =
  2. TtRR =
  3. YyPp =
  4. ttrr =
  5. ppRr =

Py, Py, Py, Py

TR, TR, tR, tR

YP, Yp, yP, yp

tr, tr, tr, tr

pR, pr, pR, pr

X X X

X X

X X X

X X

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Dihybrid Example #1

AABb AAbb AaBb Aabb

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AaBb Aabb aaBb aabb

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Axial flower position (A) is dominant to terminal flower position (a). Blue flowers (B) are dominant to white flowers (b).

  • Parent 1: Aabb
    • Gametes =
  • Parent 2: AaBb
    • Gametes =

Parent 1

Ab

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ab

Parent 2

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AB Ab aB ab

Genotype Fractions:

1/8 AABb

1/8 AAbb

2/8 = 1/4 AaBb

2/8 = 1/4 Aabb

1/8 aaBb

1/8 aabb

Genotype Ratio:

1:1:2:2:1:1

Phenotype Fractions:

3/8 Axial and Blue

3/8 Axial and White

1/8 Terminal and Blue

1/8 Terminal and White

Phenotype Ratio:

3:3:1:1

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Dihybrid Example #2

AABB AABb AaBB AaBb

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AABb AAbb AaBb Aabb

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AaBB AaBb aaBB aaBb

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AaBb Aabb aaBb aabb

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Axial flower position (A) is dominant to terminal flower position (a). Blue flowers (B) are dominant to white flowers (b).

  • Parent 1 and 2 are heterozygous for both traits
  • Parent 1: AaBb
    • Gametes =
  • Parent 2: AaBb
    • Gametes =

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Parent 1

AB

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Ab

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aB

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ab

Parent 2

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AB Ab aB ab

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R

E

S

U

L

T

s:

Genotypic Fractions:

1/16 AABB

2/16 = 1/8 AABb

1/16 AAbb,

2/16 = 1/8 AaBB

4/16 = 1/4 AaBb

2/16 = 1/8 Aabb

1/16 aaBB

2/16 = 1/8 aaBb

1/16 aabb

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AABB AABb AaBB AaBb

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AABb AAbb AaBb Aabb

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AaBB AaBb aaBB aaBb

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AaBb Aabb aaBb aabb

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Phenotypic Fractions:

9/16 Axial and Blue

3/16 Axial and White

3/16 Terminal and Blue

1/16 Terminal and White

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Parent 1

AB

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Ab

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aB

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ab

Parent 2

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AB Ab aB ab

Genotypic Ratio:

1:2:1:2:4:2:1:2:1

Phenotypic Ratio:

9:3:3:1

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Product/Multiplication Rule

The Product Rule is also known as the “AND” Rule.

  • Used when two things have to happen to get the desired outcome.
  • You will multiply the probabilities of each individual action to get the probability of the desired outcome

Basic Example: Flip a coin to get heads and roll a die to get 1

½ X ⅙ = 1/12

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Return to this Dihybrid Question

Parent 1 ‘s Genotype: Aabb and Phenotype: Axial white flowers

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Parent 2’s Genotype: AaBb and Phenotype: Axial blue Flowers

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  1. How many offspring would be axial and white?

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  • How many offspring would be terminal and white?

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Return to this Dihybrid Question

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Parent 1 and 2’s Genotypes: AaBb

Phenotype: Axial Blue Flowers

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  • How many offspring would be axial and white?

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  • How many offspring would be terminal and white?

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Relate the Product Rule to Trihybrid

TtrrAa x TTRrAa

What are the chances of having a child with the following genotypes:

    • TTrraa
    • TtRrAa
    • TTRrAA
    • TtrrAa
    • TTRRAA
    • ttrraa